So I am trying to analytically continue the formula for the partial sums of the Riemann zeta function by using the fact that, if $\zeta_{k}$ represents the $k^{\text{th}}$ partial sum of the zeta function then:

$$\zeta_{k} =\sum_{n=1}^{k} n^{-s}$$

and an alternative definition in the form of,

$$\zeta_{k}(s) = \zeta(s) - \zeta(s,k+1)$$

Both the zeta function and Hurwitz zeta function and the zeta function can be expressed in terms of Abel-Plana formulae:

$$ \begin{align} \zeta \left(s\right) &= \frac{2^{s-1}}{s-1} -2^s\int_{\mathbb{R}^{+}} \frac{\sin\left(s\tan^{-1}t\right)}{\left(1+t^2\right)^{\frac{s}{2}}\left(e^{\pi t}+1\right)} \: dt \qquad &\text{For} \: s\in \mathbb{C} \backslash \left\{1\right\}. \\ \zeta\left(s,q\right) &= \frac{p^{1-s}}{s-1} + 2\int_{\mathbb{R}^{+}} \frac{\sin\left(s\tan^{-1}\left(\frac{t}{q}\right)\right)}{\left(q^2+t^2\right)^{\frac{s}{2}}\left(e^{2\pi t}-1\right)} \: dt \: + \frac{1}{2q^s} \qquad &\text{For} \: \text{Re}\left(q\right) \geq 0. \end{align} $$

Which if we apply to the second definition of $\zeta_k$:

$$ \begin{multline} \zeta_{k}(s)=\frac{2^{s-1}-(k+1)^{s-1}}{s-1}-\frac{1}{2(k+1)^s} \\ -2\int_{\mathbb{R}^+} \frac{2^{s-1}\sin{\left(s\tan^{-1}t\right)}}{(1+t^2)^{s/2}(e^{\pi t}+1)} + \frac{\sin{\left(s\tan^{-1}\left(\frac{t}{k+1}\right)\right)}}{((k+1)^2+t^2)^{s/2}(e^{2\pi t}-1)} \: dt \end{multline} $$

By analytic continuation the formula for the partial sums of the zeta function has been extending from the positive integers to allowing complex arguments.

Is this analytic continuation correct? Also are there any similar analytic continuations involving trig ratios? Any input would be greatly appreciated

  • 1
    $\begingroup$ $\sum_{n=1}^k n^{-s}$ is already entire $\endgroup$ – reuns Aug 14 '16 at 3:20
  • $\begingroup$ Just want to point out that this generalises the definition of the partial sum of the zeta function as k can be any real number except for -1 as that produces a singularity. $\endgroup$ – 193406573160514765gdhssg Jun 3 at 14:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.